Maximum XOR Queries - Google Top Interview Questions
Problem Statement :
You are given a list of non-negative integers nums and a two-dimensional list of integers queries where each element contains [x, limit]. Return a list such that for each query [x, limit], we find an e in nums such that e ≤ limit and XOR(e, x) is maximized. If there's no such e, use -1. Constraints n ≤ 100,000 where n is the length of nums m ≤ 100,000 where m is the length of queries Example 1 Input nums = [2, 4, 8] queries = [ [3, 5], [2, 0] ] Output [4, -1] Explanation For the first query, we can use 2 or 4 in nums. 2 ^ 3 = 1 while 4 ^ 3 = 7 so we pick 4 which yields the bigger XOR. In the second query, there's no number that's less than or equal to 0, so we set it to -1.
Solution :
Solution in C++ :
struct TrieNode {
int lo = INT_MAX;
TrieNode* children[2]{};
};
class Solution {
public:
vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
TrieNode* root = new TrieNode();
for (int num : nums) {
TrieNode* p = root;
for (int i = 30; i >= 0; --i) {
int nxt = (num & (1 << i)) ? 1 : 0;
if (!p->children[nxt]) p->children[nxt] = new TrieNode();
p = p->children[nxt];
p->lo = min(p->lo, num);
}
}
vector<int> ret;
for (auto q : queries) {
int x = q[0], limit = q[1];
int ans = 0;
TrieNode* p = root;
for (int i = 30; i >= 0; --i) {
if (x & (1 << i)) {
if (p->children[0]) {
p = p->children[0];
} else if (!p->children[1] || (p->children[1]->lo > limit)) {
ret.emplace_back(-1);
break;
} else {
p = p->children[1];
ans ^= (1 << i);
}
} else {
if (p->children[1] && (p->children[1]->lo <= limit)) {
p = p->children[1];
ans ^= (1 << i);
} else if (!p->children[0]) {
ret.emplace_back(-1);
break;
} else {
p = p->children[0];
}
}
if (i == 0) ret.emplace_back(ans);
}
}
return ret;
}
};
vector<int> solve(vector<int>& nums, vector<vector<int>>& queries) {
return (new Solution())->solve(nums, queries);
}
Solution in Java :
import java.util.*;
class Solution {
public int[] solve(int[] nums, int[][] qs) {
int n = nums.length;
List<Map<Integer, Integer>> ls = new ArrayList<>();
for (int i = 0; i < 32; i++) {
Map<Integer, Integer> mp = new HashMap<>();
for (int x : nums) {
// set.add(x >> i);
mp.put(x >> i, Math.min(x, mp.getOrDefault(x >> i, Integer.MAX_VALUE)));
}
ls.add(mp);
}
Arrays.sort(nums);
int[] res = new int[qs.length];
Arrays.fill(res, -1);
if (n == 0)
return res;
int cnt = 0;
for (int[] q : qs) {
int v = q[0];
int limit = q[1];
int ans = 0;
boolean find = false;
for (int i = 31; i >= 0; i--) {
Set<Integer> ts = ls.get(i).keySet();
Map<Integer, Integer> mp = ls.get(i);
// if(i == 0)System.out.println(ts);
if ((v | (1 << i)) == v) { // this is one
int tk = ans >> i;
if (!ts.contains(tk)) {
ans |= 1 << i;
}
// System.out.println(tk+" "+i+" "+ts+" "+ans);
} else { // we want to get index to be 1
int now = ans | (1 << i);
int tk = now >> i;
/// if we need this index but no that index
if (ts.contains(tk) && mp.get(tk) <= limit) {
ans = now;
}
// System.out.println(tk+" "+i+" "+ts+" "+ans);
}
}
if (ans <= limit && ans >= nums[0]) {
res[cnt] = ans;
}
cnt++;
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums, queries):
nums.sort()
def query(x, limit):
start = 0
stop = bisect_right(nums, limit)
num = 0
for i in range(31, -1, -1):
i_th_bit = 1 << i
plus = num + i_th_bit
if nums[start] >= plus:
num = plus
elif nums[stop - 1] >= plus:
cut = bisect_left(nums, plus, start, stop)
if x & i_th_bit:
stop = cut
else:
start = cut
num = plus
return num
return [query(x, limit) if nums and nums[0] <= limit else -1 for x, limit in queries]
View More Similar Problems
Left Rotation
A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d
View Solution →Sparse Arrays
There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun
View Solution →Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →